SOLUTIONS Section (a) y' + 4 x+2 y = -6. (x+2) 2, P(x) = 4, P(x) dx = 4 applen(x+2) = applen(x+2)4. = (x+2) 4,

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1 page of Section 4. solutions SOLUTIONS Section 4.. (a) y' + 4 x+2 y = -6 (x+2) 2, P(x) = 4, P(x) dx = 4 applen(x+2) = applen(x+2)4 x+2 applen(x+2) 4 I = e = (x+2) 4, answer is y = -2 x+2 + k (x+2) 4 (x+2) 4 y = -6(x+2) 2 dx = -6(x+2)3 3 + k, (b) y' + 4xy = x, P(x) = 4x, P(x) dx = 2x 2, I = e 2x2, ye 2x2 = xe 2x2 dx = 4 e2x2 + K. Answer is y = 4 + Ke-2x2 (c) method Rewrite the equation as y' + y = 2 e2x. Then P(x) =, I = e x, ye x = e x 2 e2x dx= 2 e3x dx = 6 e3x + K. Answer is y = 6 e2x + Ke -x. method 2 (since the coeffs are constant) (no need to rewrite the equation) m = -, y h = Ae -x. Try y p = Be 2x. Then 4Be 2x + 2Be 2x = e 2x, 6B =, B = 6 y p = 6 e2x, answer is y = Ae -x + 6 e2x (d) y' - y cot x = csc x, P(x) = - cot x, P(x) dx = -applen sin x = applen csc x, I = e applen csc x = csc x, y csc x = csc 2 x dx = - cot x + K Answer is y = - cos x + K sin x 2. Rewrite as x(y')' + y' = 4x, (y')' + x y' = 4. Then P(x) dx = applen x, I = e applen x = x, xy' = 4x dx = 2x 2 + K so y' = 2x + K. Antidiff to get x y = x 2 + K applen x + C (two arbitrary constants) 3. (a) y' + 4xy = x P(x) = 4x, Q(x) = x P(x) dx = x 2 + K I = e x2 +K = e x 2 e K which you could call Ae x2 but I'll just leave it e x2 e K (doesn't make any difference) Iy = IQ e x2 e K y = e x2 e K x dx = e K ex2 x dx The e K on both sides cancels out now and you're left with e x2 y = e x2 x dx etc. Any K you put in at the P(x) dx stage will cancel out later so why bother. The only constant you'll have left at the end of the problem is the one you put in when you find IQ.

2 page 2 of Section 4. solutions (b) You get solution y = /4. This is a solution but the general solution. If you have an IC to satisfy, unless you are very very lucky, your one particular solution won't satisfy that IC and you have no constants available to make it satisfy the IC. 4. method y' - ky = 0, m = k, y = Ae kx method 2 y' - ky = 0, P(x) = -k, P(x) dx = -kx, I = e -kx, ye -kx = 0 dx = C, y = Ce kx 5. method P(x) =, P(x) = x, I = e x, ye x = dx = x + K, y = xe -x + Ke -x To get y(-) = 3 need 3 = -e + Ke, K = 3 + e e Answer is y = xe -x e e -x = xe -x + 3e -x- + e -x e method 2 (since coeffs are constant) m = -, y h = Ae -x. Try y = Bxe -x (step up). Need p -Bxe -x + Be -x + Bxe -x = e -x xe -x terms drop out. Equate coeffs of the e -x terms B = Gen sol is y = Ae -x + xe -x. The IC determine A as in method. 6. (a) e K is not quite arbitrary. It can never be zero and it can never be negative. So it really shouldn't be turned into a C which is totally arbitrary. But most people pay no attention to these niceties. And it usually works out OK in the long run. You would probably find that the new not-so-arbitrary constant C (that used to be e K ) comes out positive anyway when you plug in a realistic condition. (b) Yes because applen K can take on any value, from very negative to very positive. On the other hand, you have a slight problem because you shouldn't be taking log of an arbitrary K because you can't take applen of a negative number or zero. Most people don't worry about this either. 7. y' + 2 x y = x2 + x. Then P(x) = 2 x, P(x) = 2 applen x = applen x2, I = x 2, x 2 y = x(x 2 + ) dx = (x 3 + x) dx = x4 4 + x2 2 + K, y = x K x 2 8. First solve y' - x y = x. P(x) = - x, P(x) = -applen x = applen x-, I = x, x y = dx = x + C, y = x2 + Cx. Then solve y' - x y = 0 getting y = 0 dx = K, y = Kx. So x

3 page 3 of Section 4. solutions y = x 2 + Cx if x 3 Kx if x > 3 The condition y() = 2 makes C =. To get the sol continuous, we want x 2 + x = Kx when x = 3, 2 = 3K, K = 4 Answer is y = x 2 + x if x < 3 4x if x > 3

4 page 4 of Section 4. solutions HONORS 9.(a) The equation is y' - ry = -h where r and h are (positive) constants. It's a linear first order DE with constant coeffs. method for solving the DE m = r, y h = Ae rt Try y p = B. Need 0 - rb = -h, B = h/r y gen = Ae rt + h r method 2 for solving the DE P = -r, Q = -h, I = e -rt e -rt y = -h e -rt dt = h r e-rt + A y = h r + Aert Whichever method you used, plug in the IC y(0) = N to get A = N - h r Solution is y(t) = h r + (N - h r )ert (b) The solution now is y(t) = 2 h + (40-2 h)e2t Fished out means that as t gets larger, y hits 0 eventually. So the problem is to find which h's let y(t) reach 0. Here's the graph point of view. The graph of y(t) = 2 h + (40-2 h)e2t starts at height 40 when t = 0. It's decreasing if 40-2 h < 0 h > 80 and it's increasing if h < 80. So the lake gets fished out if h > 80. To illustrate, here's the graph of y(t) for h = 60 (and you can see the fish population taking off) and again for h = 90 where you see the population hit 0 (at which point the mathematical model ceases to apply.)

5 page 5 of Section 4. solutions Plot[60/2 + (40-60/2)E^(2t),{t,0,2}]; 40 2 Plot[00/2 + (40-00/2)E^(2t),{t,0,.5}, Ticks-{{.5,,.5},{20,40}}]; (c) ( )e2t = 0, e 2t = 5, t = 2 applen 5 In the second diagram, the graph crosses the horizontal axis at t = applen 5. 2 That's when the lake is fished out.

6 page of Section 4.2 solutions SOLUTIONS Section 4.2. (a) cos y dy = -x dx, sin y = - 2 x2 + C (implicit sol) (b) y dy = - dx x 3, 2 y2 = 2x 2 + C, y = x 2 + D (c) y 4 dy = -x 2 dx, 5 5 y5 = - 3 x3 + C, y = K x3 (d) dy y = dx 2x + 3, applen Ky = 2 applen(2x + 3) = applen 2x + 3, Ky = 2x + 3, y = A 2x + 3 warning It's OK to write applen y = 2 applen(2x + 3) + C but when you take exp on both sides it is wrong to get y = 2x + 3 PLUS e C WRONG which turns into y = 2x A The right way to take exp is to get applen(2x+3) /2 + C y = e RIGHT which turns into /2 C y = eapplen(2x+3) TIMES e = A 2x + 3 (e) e -y dy = dx x 2, - e-y = - x + C, e-y = x + D, -y = applen( x + D), y = - applen( x + D) (f) y dy = (5x + 3) dx, 2 y2 = 5 2 x2 + 3x + C, y = 5x 2 + 6x + D (g) not separable (h) (y + ) = x dx, 2 y2 + y = applen Kx (implicit sol), y = applenkx 2 = - = 2 apple Kx (explicit sol)

7 page 2 of Section 4.2 solutions 2. For (e), y = - applen ( x + D) x 2 dy = x 2 y'(x) dx = x 2 - x + D - x 2 dx = dx x + D e -y dx = e-applen ( x + D) applen ( x + D)- = e dx = ( x + D)- dx So x 2 dy does equal e -y dx, QED. For (f), y = 5x 2 + 6x + D, y' = so y' does equal 5x + 3, QED. y 0x x 2 + 6x + D = 5x + 3 5x 2 + 6x + D 3.(a) dy y = x dx, applen Ky = x 2 /2 x 2 /2 2 x2, Ky = e, y = Ae Use the condition to get 3 = Ae /2, A = 3e -/2, -/2 Sol is y = 3e (b) y dy = (3-5x)dx, Then ex 2 (x 2 -)/2, y = 3e 2 y2 = 3x x2 + C. Set x = 2, y = 4 to get C = 2. 2 y2 = 3x x2 + 2, y = 6x - 5x (Choose the positive square root since y is positive when x = 2.) (c) e y dy = 3x dx, e y = 3 2 x2 + C. Set x = 0, y = 2 to get C = e 2. Then e y = 3 2 x2 + e 2, y = applen ( 3 2 x2 + e 2 ) (d) dy = cos x dx, - y4 3y 3 = sin x + C. Set x = 0, y = 2 to get C = Sol is y 3 3 sin x method 2 (works because the coeffs are constant) w' + 5 w = 2, m = - 5, w h = Ae-t/5 Try w p = B. Substitute into the DE to get B = 2, B = 0 w gen = Ae -t/5 + 0 (which is the same as the answer 0 - De -t/5 from example because the arbitrary constant A is the same as the arbitrary constant -D) method 2 P = 5, Q = 2, I = e P dt = e t/5 Iw = IQ = 2et/5 dt, e t/5 w = 0e t/5 + K, w = 0 + Ke -t/5 5. The equation can be written as xy 2 - y - 7 = 0 and treated as a quadratic equation of the form ay 2 + by + c = 0 where a = x, b = -, c = -7. So y = +28x, not really one implicit solution, but two. (And they real only if 2x x > -/28 and they aren't defined for x = 0.)

8 page 3 of Section 4.2 solutions 6.(a) dy y = -0 dx, applen Ky = -0x, Ky = e-0x, y = Ae -0x. Plug in the IC y(0) = G to get y = Ge -0x. Now let y = G/2 and find x: G/2 = Ge -0x, 2 = e-0x, -0x = applen /2, x = 0 applen 2. So the half life is 0 applen 2. (b) Let the constant of proportionality be called C. As in part (a), the half life is C applen 2.If you want this to be 3, choose C = 3 applen (a) Differentiate w.r.t. x to get the differential equation of the family xy = K. xy' + y = 0 y' = -y/x The orthog family has differential equation y' = x/y. Solve it to find the orthogonal family. y dy = x dx 2 y2 = 2 x2 + A y 2 - x 2 = C Both families are hyperbolas (each hyperbola has two branches). In the diagram, the original family is in darker type. (b) y x 2 = A x -2 y' - 2x -3 y = 0 y' = 2y x The orthog family has DE y' = - x. Solve to get the orthog family. 2y 2yy' = -x 2y dy = -x dx y 2 = - 2 x2 + C x 2 + 2y 2 = D (a family of ellipses) K=- K=2 C=-/2 C= A=3 A=/2 C=3 D=5 D=9 D= A=-4 Problem 7(a) Problem 7(b)

9 page of Section 4.3 solutions SOLUTIONS Section 4.3. d( y ) comes out to be (22) immediately by the quotient rule and similarly for (23) x d(x 2 + y 2 ) - = -(x 2 + x 2 ) -2 d(x 2 + y 2 )) (chain rule) = -(x 2 + x 2 ) -2 (2x dx + 2y dy) which is (24) d( x 2 + x 2 ) = 2 (x2 + y 2 ) -/2 d(x 2 + y 2 ) (chain rule) = 2x dx + 2y dy which cancels to (25) 2 x 2 + x 2 d(applen (x 2 + y 2 ) = d(x2 + y 2 ) (chain rule) = x 2 + y 2 2x dx + 2y dy (x 2 + y 2 ) d(arctan y/x) = + (y/x) 2 d(y x ) = x dy - y dx + (y/x) 2 x 2 = -y dx + x dy x 2 + y 2 2. x = r cos œ so by (), dx = cos œ dr - r sin œ dœ y = r sin œ, dy = sin œ dr + r cos œ dœ 3.(a) p = 2xy, q = y, ºq ºx = 0, ºp = 2x. Not equal so form is not exact ºy (b) ºq ºx = ºp ºy = 3x2. Exact. Antidiff p w.r.t. x to get 4 x4 + x 3 y. Diff this temporary answer w.r.t. y to get x 3. Compare with q and see that you should tack on 4 y4. Answer is f(x,y) = 4 x4 + x 3 y + 4 y4 (c) f(x,y) = - y x + 5y 4. Need ºq ºx = ºp ºy, ºq ºx = 3xy2, q = 3 2 x2 y 2 + any f(y) For example q could be 3 2 x2 y 2 + sin y (a) d(2x 3 + xy 2 + y 3 ) = 0, implicit sol is 2x 3 + xy 2 + y 3 = C (b) d(x 3 + xy) = 0 Implicit sol is x 3 + xy = C. Explicit sol is y = C - x3 x (c) (x-y cos x) dx - (y + sin x) dy = 0 d( 2 x2 - y sin x - 2 y2 ) = 0 Implicit solution is 2 x2 - y sin x - 2 y2 = K (d) e xy dx - dy= 0 Not exact since ºq ºx (e) (2r cos œ - )dr - r 2 sin œ dœ = 0 d(r 2 cos œ - r) = 0 Implicit sol is r 2 cos œ - r = K = 0 but ºp ºy = xexy (f) not exact (g) d(sin x cos y) = d( 4 x4 ) Implicit sol is sin x cos y = 4 x4 + C

10 page 2 of Section 4.3 solutions (h) (ye -x - sin x) dx - (e -x + 2y) dy = 0 d(-ye -x + cos x - y 2 ) = 0 Implicit sol is -ye -x + cos x - y 2 = C 6. Take differentials throughout -ye -x + cos x - y 2 = C to get -y e -x dx + e -x -dy - sin x dx - 2y dy = 0 Collect terms: (ye -x - sin x) dx = (e -x + 2y) dy, QED 7.(a) d(x 2 y + 2 y2 ) = 0 Implicit sol is x 2 y + 2 y2 = C Set x =, y = 4 to get C = 2. Implicit particular sol is is x 2 y + 2 y2 = 2 (b) d(-cos(2x + 3y) ) = 0 Implicit sol is -cos(2x + 3y) = C Set x = 0, y = π/2 to get C = 0. Implicit particular sol is cos(2x + 3y) = 0 (c) d applen(x + y) = d(x). Implicit sol is applen(x + y) = x + C. Set x = 0, y = to get C = 0. Implicit particular sol is applen(x + y) = x. Then x + y = e x so explicit solution is y = e x - x 8.(a) d( 3 x3 + 2x y2 ) = 0 Implicit sol is 3 x3 + 2x y2 = K (b) (x 2 + 2)dx = -3y dy, 3 x3 + 2x = y2 + K 9. (a) It doesn't do any good to collect the dx terms and get (x 2 + y 2 + y) dx - x dy = 0 because this arrangement isn't exact. Instead, look at (27). Use integrating factor x 2. The equation becomes + y2 x dy - y dx dx = x 2 + y 2 so x = tan - y x + K (b) See (22). Use integrating factor /y 2. y dx - x dy y 2 = dx, (c) See (25). Use integrating factor x x y = x + K, y = x + K. x 2 + y 2 dy = x dx + y dy, y = x 2 + y 2 + K x 2 + y 2 (d) x dx = (x 2 + y 2 - y) dy x dx + y dy = (x 2 + y 2 )dy See (26). Use integrating factor 2 x 2 + y 2.

11 page 3 of Section 4.3 solutions 2x dx + 2y dy = 2 dy, applen(x 2 + y 2 ) = 2y + K x 2 + y 2 (e) See (23). Multiply by /x 2 to get x dy - y dx x 2 = 2x dx + 2y dy, d( y x ) = d(x2 + y 2 ), y x = x2 + y 2 + K. Implicit sol is y = x 3 + xy 2 + Kx.

12 page of Section 4.4 sols SOLUTIONS Section 4.4. Remember that at each point (x,y), the idea is to draw a little segment with slope x/y. On the line y = x, all the segments have slope. In quadrants I and III the segments all have positive slope. In quadrants II and IV the segments have negative slope As you move right on a horizontal line in quadrants I, y stays the same and x gets larger so the segments get steeper. As you move left on a horizontal line in quadrant II, y stays the same and x gets negatively larger so the segments get steeper (but with negative slope). As you move up a vertical line in quadrant I, x stays the same and y gets larger so the segments get less steep. etc. MyField = directionfield[x/y, {x,-3,3},{y,-3,3},.5,.3]; Show[MyField, Axes->True, Ticks->None]; y dy = x dx y 2 2 = x2 2 + K y 2 - x 2 = A Each solution is a hyperbola. Here's a picture of the direction field along with the two particular solutions y 2 - x 2 = 2, y 2 - x 2 = -2 SomeSols = ImplicitPlot[{y^2 - x^2== -2,y^2 - x^2 ==2},{x,-3,3}, PlotStyle->{{GrayLevel[.5],Thickness[.0]}}, PlotRange->{-3,3},DisplayFunction->Identity]; Show[{MyField,SomeSols}, Axes->True, Ticks->None,DisplayFunction->$DisplayFunction];

13 page 2 of Section 4.4 sols (b) When y = x, the little segments have slope. As you move right on a horizontal line in quadrant I, y stays fixed and x increases so the segments get less steep. As you move up a vertical line in quadrant I, x stays fixed and y increases so the segments get steeper. dy y = dx x applen Ky = applen x Ky = x y = Ax The solutions are all lines through the origin. 2. Draw a curve through the point (0,) using the little segments. (By the way, the direction field in the problem happens to be that of the DE y' = xy.) 3.(a). The segment at point (x, y ) is supposed to have slope x y. If you solved the DE and found the particular solution satisfying the condition y(x ) = y, in the 0 0 vicinity of point (x, y ) it's graph should look like the little segment. 0 0 (b) The equation is separable and linear first order and exact. I'll separate. dy y = x dx applen Ky = 2 x2 Ky = e x2 /2 y = Ae x2 /2

14 page 3 of Section 4.4 sols (c) I plotted 2e x2 /2, 4e x 2 /2 and -4e x 2 /2. A=4 A=2 A=-4 (d) Plug in the condition y(4) = 3; i.e., set x=4, y=3 to determine A. 3 = Ae 8, A = 3e -8. The curve is y = 3e -8 e x2 /2 ; equivalently y = 3e -8+x 2 /2.

15 page of solutions to review problems for Chapter 4 SOLUTIONS review problems for Chapter 4.(a) method (exact) d( 3 x3 + 2x y2 )= 0, 3 x3 + 2x y2 = K 2 y = 3 K x3-4 3 x method 2 (separable) (x 2 + 2) dx = -3y dy, 3 x3 + 2x = y2 + K, same y as in method now. (b) method (separable) dy y = - dx, applen Ky = -x, Ky = e-x, y = Ae -x method 2 (exact) y dx + dy = 0 is NOT exact but dx + y dy = 0 is exact. Then d(x + applen y) = 0, x + applen y = K, applen y = K - x, y = e K-x, y = e K e -x, y = Ae -x method 3 (linear, constant coeffs) y' + y = 0, m = -, y = Ae -x method 4 (linear first order) y' + y = 0, P =, Q = 0, I = e P = e x, e x y = 0 dx = K, y = Ke -x (c) method (can be arranged to be exact) (y-2x) dx + x dy = 0, d(xy - x 2 ) = 0, xy - x 2 = C If x =, y = 2 then C =. Sol is xy - x 2 =. Explicit sol is y = x + x method 2 (first order linear) y' + x y = 2, I = e (/x) = e applen x = x, xy = 2x dx = x 2 + K, y = x + K x. If y() = 2 then K =. Sol is y = x + x (d) (linear) y'' - y = 0, m =, y = Ae x + Be -x (e) method y'' - 3y' = 2, m = 0,3, y h = A + Be 3x Try y p = Cx (step up) Get C = -4. Answer is y = A + Be 3x - 4x method 2 If you think of y' as the variable this is first order. Let y' = u Then DE is u' = 3u + 2 This is exact, also separable,. also linear with constant coeffs (P,Q stuff). Here's the separation method: du 3u + 2 = dx, applen K(3u + 2) = x, 3 applen K(3u + 2) = 3x, K(3u + 2) = e 3x u = Be 3x - 4 So y' = Be 3x - 4 and (antidiff) y = 3 Be3x - 4x + C, y = De 3x - 4x + C, same as in other method

16 page 2 of solutions to review problems for Chapter 4 (f) method (separable) dy dx = ex e y, e -y dy = e x dx, -e -y = e x + K, e -y = A - e x, -y = applen(a - e x ), y = - applen(a - e x ) method 2 (exact) e x dx - e -y dy = 0, d(e x + e -y ) = 0, e x + e -y = K, y = - applen(k - e x ) (g) (Second order, variable coeffs, can't use m's) method Consider y' the variable (call it w if you like). Then the equation is first order separable: dw xw' - w =, w+ = dx, applen(w + ) = applen Kx, w + = Kx, w = Kx - x So y' = Kx -, y = Kx 2 - x + C The IC make K = 2, C = Answer is y= 2x 2 - x + method 2 Consider y' the variable (call it w if you like). Then the equation is first order linear: w' - x w = x P(x) I = e dx -applen = e x applen /x = e = x x w = x x dx = - x + K. So y' = - + Kx, y = -x + 2 Kx2 + C. The IC make K = 4, C =, Answer is y = -x + 2x method (separable) m dv = mg - cv dt dv vc - mg = - dt m c applen(cv - mg) = - t m K(cv - mg) = e -ct/m cv - mg = A e -ct/m v = mg c + A c e-ct/m method 2 Use the P,Q method. v' + c m v = g, I = e c/m ex = e ct/m, e ct/m y = ge ct/m dt = gm c ect/m + K, y = gm c + Ke-ct/m etc. method 3 (harder to notice) m dv + (cv - mg) dt = 0 is not exact but m cv - mg p dv + dt = 0 q

17 page 3 of solutions to review problems for Chapter 4 is exact (ºq/ºv = ºp/ºt = 0). The equation can be written as d( m c applen(cv - mg) + t) = 0 Implicit solution is m c applen(cv - mg) + t = K Solve for v to get explicit sol: applen(cv - mg) = Kc m - ct m cv - mg = e Kc/m - ct/m = e Kc/m e -ct/m = A e -ct/m v = mg c + A c e-ct/m 3. At point (x,y) draw a little segment with slope x+y. method for solving The equation is y' - y = x. This is linear first order with constant coeffs so the methods of Chapter work. m =, y h = Ae x. Try y p = Bx + C. Substitute into the DE. You need B - (Bx + C) = x. Equate the x coeffs: -B = c, B = -. Equate the constant terms: B - C = 0, C = -. So y p = -x-, y gen = y h + y p = Ae x - x - x. method 2 for solving The equation is linear. It can be written as y' - y = x. P(x) = -, Q(x) = x, e P(x)dx = e -x, ye -x = xe-x dx = -xe -x - e -x + C. Solution is y = -x - + Ce x. Yes you should be able to sketch the graph of this family. I would first sketch y=-x- (a line) and y = Ce x separately.

18 page 4 of solutions to review problems for Chapter 4 y=0e x y = 2 ex y = - 2 ex y=-x- y=-0e x Then add the line heights to each of the other curves. Way out to the left, Ce x is near 0 so the sum is like the line. Way out to the right, the exponential heights are much larger in absolute value than the line heights so the sum is like the exponential. When C = 0, the sum is the line. Here are some of the curves in the family y = -x - + e x, along with the direction field. C=2 C=0 C=-2

19 page 5 of solutions to review problems for Chapter 4 4. Rewrite the DE as y = A and then differentiate w.r.t. x on both sides. x3 x -3 y' -3x -4 y = 0 y' = 3y x The orthogonal family has differential equation y' = - x. Now solve it. 3y 3y dy = - x dx 3y 2 2 = - x2 2 + C x 2 + 3y 2 = 2C x 2 + 3y 2 = K (an ellipse family) A=/2 K=9 K=5 K= A=-4 5. A separable DE can be written as ( ) x stuff dx = y stuff dy and rearranged to look like ( ) x stuff dx + -y stuff dy = 0 p q Then ºq ºx = ºp ºy (both are 0) so ( ) is exact. To disprove the converse, i.e., to show that not every exact DE is also separable, all you have to do is produce one counterexample. Problem (c) is exact but not separable. QED